Solve the simultaneous equation, 5x=2y+1 2x-3y+4+0

Answers

Answer 1

Answer: Hello,

I suppose the 2th equation is 2x-3y+4=0

5x-2y=1 ........|*2 | *3
2x-3y=-4....... |*(-5) | *-2
==> -4y+15y=2+20==>11y=22==>y=2
and 15x-4x=3+8 ==>11x=11 ==>x=1
Proof:

5*1=? 2*2+1 yes
2*1-3*2+4=? 0 yes (linear combinaison method)

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Don randomly draws two cards from a standard deck of 52 cards. He does not replace the first card. What is the probability that both cards are aces?

Answers

easy
so remember
4 aces in a deck
probability =(desired outcomes)/(total possible)

total possible=52

4/52=1st card
then since he picked 1 out the total possible is and assuming he drwe an ace 3/51
so the answer is 4/52 times 3/51=1/221

Answer: 1/221

Step-by-step explanation:

A P E X

I need help please

Answers

Answer:

help on what like whats ur question

Step-by-step explanation:

Answer:

I hope this helps

Step-by-step explanation:

need more letters

Sins = 8x2 and = 2x + a find 5- o

Answers

(f × g ) (w) (Please note that I have used w in place of x)
∴((8w²) × (2w+8) × w
(16w³ × 64w²) × w
$16w^(4)$ × 64w³
$1024w^(7)$

Which long division problem can be used to prove the formula for factoring the difference of two perfect cubes?

Answers

Some of the possible options of the questions are;

A) $(a - b) | \overline {a^2 + a \cdot b + b^2}$

B) $(a + b) | \overline {a^2 - a \cdot b + b^2}$

C) $(a + b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

D) $(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

The difference of two perfect cubes has a binomial factor and a trinomial factor

The option that gives the long division problem that can be used to prove the difference of two perfect cubes is option D

D) $\underline {(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}}$

Reason:

The formula for factoring the difference of twoperfect cubes is presented as follows;

a³ - b³ = (a - b)·(a² + a·b + b²)

Given that a factor of the difference of two cubes is (a - b), and that we

have; (a³ + 0·a·b² + 0·a²·b - b³) = (a³ - b³), both of which are present in

option D, by long division of option D, we have;

${} \hspace {33} a^2 + a \cdot b + b^2\n(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a^2 \cdot b - b^3}\n{} \hspace {33} \underline{a^3 - a^2 \cdot b }\n{} \hspace {55} a^2 \cdot b + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3\n {} \hspace {55} \underline{a^2 \cdot b - a \cdot b^2}\n{} \hspace {89} a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3\n{} \hspace {89} \underline{a \cdot b^2 - b^3}\n{}\hspace {89} 0$

By the above long division, we have;

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ = a² + a·b + b²

Which gives;

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ = (a³ + 0·a·b² + 0·a·b² - b³)/(a - b)

We get;

(a³ + 0·a·b² + 0·a·b² - b³)/(a - b) = a² + a·b + b²

(a - b)·(a² + a·b + b²) = (a³ + 0·a·b² + 0·a·b² - b³) = (a³ - b³)

(a - b)·(a² + a·b + b²) = (a³ - b³)

(a³ - b³) = (a - b)·(a² + a·b + b²)

Therefore;

The long division problem that can be used to prove the formula for

factoring the difference of two perfect cubes is

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ , which is option D

D) $(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

Learn more here:

brainly.com/question/17022755

Answer:

The correct options, rearranged, are:

Options:

$A)(a^2+ab+b^2)/(a-b)\n\nB)(a^2-ab+b^2)/(a+b)\n\nC)(a^3+0a^2+0ab^2-b^3)/(a+b))\n\n D)(a^3+0a^2+0ab^2-b^3)/(a-b)$

And the asnwer is the last option (D).

Explanation:

You need to find which long division can be used to prove the formula for factoring the difference of two perfect cubes.

The difference of two perfect cubes may be represented by:

$a^3-b^3$

And it is, as a very well known special case:

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Then, to prove, it you must divide the left side, $a^3-b^3$ , by the first factor of the right side, $a-b$

Note that, to preserve the places of each term, you can write:

$(a^3-b^3)=(a^3+0a^2+0ab^2-b^3)$

Then, you have:

$(a^3+0a^2+0ab^2-b^3)=(a-b)(a^2+ab+b^2)$

By the division property of equality, you can divide both sides by the same factor, which in this case will be the binomial, and you get:

$(a^3+0a^2+0ab^2-b^3)/(a-b)=(a^2+ab+b^2)$

That is the last option (D).

Which comes first domain or range

Answers

they are intertwined the domain is the allowed values of the independent varialbles and the range is the allowed values of the dependent variable.

in order for this to work.range comes first because for example a 100-500 range between the phone cost

Solve:

3/5 − 6/11

A:3/55

B:3/6

C:3/11

Answers

The answer is: [A]: 3/55 .
__________________________________
Explanation:
__________________________________
(3/5) - (6/11) = [(3*11) / (5*11) ] - [(6*5) / (11*5)] = (33/55) - (30/55) = 3/ 55.
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